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ololo11 [35]
3 years ago
6

In class I mentioned that we used to show an actual ballistic pendulum where a rifle is shot into a heavy block to see how the b

lock moves. Assume a 4.7 gram bullet is fired at 678 m/s into a 4.8 kg block of wood. The block hangs from a long string to approximate 1D motion. How fast does the block move just after the collision in m/s?
Physics
1 answer:
Len [333]3 years ago
3 0

Answer:

0.66 m/s

Explanation:

This is an inelastic collision, for that reason the conservation of momentum law allows us to determine the final velocity of both, the bullet and block together after the collision:

m_{bullet} v_{bullet0} +m_{block} v_{block} =(m_{bullet}+m_{block})v_{f} \\v_{f} =\frac{m_{bullet} v_{bullet0} }{m_{bullet}+m_{block}} \\v_{f} =\frac{0.0047kg*678m/s}{4.8kg+0.0047kg}\\v_{f} =0.66 m/s

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Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

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From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

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F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

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F = 0 N

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2 years ago
A 63 gg ice cube can slide without friction up and down a 30∘30∘ slope. The ice cube is pressed against a spring at the bottom o
tatiyna

Given Information:

slope angle = θ = 30°

spring constant = k = 30 N/m

compressed length = x = 10 cm = 0.10 m

mass of ice cube = m = 63 g = 0.063 kg

Required Information:

distance traveled by ice cube = d = ?  

Answer:

distance traveled by ice cube = 0.48 m

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Using the the principle of conversation of energy, the following relation holds true for this case,

mgh = 1/2*kx²

h = 1/2*kx²/mg

Where h is the height of the slope, m is the mass of ice cube, k is the spring constant and x is the compressed length o the spring and g is gravitational acceleration.

h = 1/2*kx²/mg

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sinθ = h/d

d = h/sinθ

d = 0.242/sin(30)

d = 0.48 m

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